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<meta name="description" content="《programming for the puzzled》第11章涉及到的算法和语言问题：理解列表，递归进行分治搜索。有2^n×2^n大小的院子，要用面积为3的L形的瓷砖铺满可以在不突破边界，不破坏瓷砖，瓷砖之间也不重叠的情况下完成吗？答案是不行，因为2^n×2^n不能被3整除，只能被2整除。而如果允许有一个方块的地方可以被剩下来，则2^2n-1则可以被3整除。有没有一个铺砖的算法能够将任意2^n">
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          <h1 class="post-title" itemprop="name headline">量化投资学习笔记57——通过问题学算法11:请给院子铺瓷砖（Tile That Courtyard, Please）</h1>
        

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        <p>《programming for the puzzled》第11章<br>涉及到的算法和语言问题：理解列表，递归进行分治搜索。<br>有2^n×2^n大小的院子，要用面积为3的L形的瓷砖铺满<br><img src="https://zymblog-1258069789.cos.ap-chengdu.myqcloud.com/blog0178-QTLearn/39/01.png"><br>可以在不突破边界，不破坏瓷砖，瓷砖之间也不重叠的情况下完成吗？答案是不行，因为2^n×2^n不能被3整除，只能被2整除。而如果允许有一个方块的地方可以被剩下来，则2^2n-1则可以被3整除。<br>有没有一个铺砖的算法能够将任意2^n×2^n大小的院子用面积为3的L形的砖铺满只剩一个方块没铺？<br>归并排序是分治算法的一个例子。<br>要排序一个序列（如[a, b, c, d]），可以把这个序列分成两个序列([a,b], [c,d])，然后递归地对两个序列进行排序。如果序列长度为2，则比较这两个元素，结束递归。最后用一个合并的函数将两个列表合并到一起。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># coding:utf-8</span></span><br><span class="line"><span class="comment"># 《programming for the puzzled》实操</span></span><br><span class="line"><span class="comment"># 11.瓷砖铺地</span></span><br><span class="line"></span><br><span class="line"></span><br><span class="line"><span class="comment"># 分治法，归并排序</span></span><br><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">mergeSort</span>(<span class="params">L</span>):</span></span><br><span class="line">    <span class="comment"># 不加下面这行递归无法结束</span></span><br><span class="line">    <span class="keyword">if</span> <span class="built_in">len</span>(L) &lt; <span class="number">2</span>:</span><br><span class="line">        <span class="keyword">return</span> L</span><br><span class="line">    <span class="keyword">if</span> <span class="built_in">len</span>(L) == <span class="number">2</span>:</span><br><span class="line">        <span class="keyword">if</span> L[<span class="number">0</span>] &lt;= L[<span class="number">1</span>]:</span><br><span class="line">            <span class="keyword">return</span> [L[<span class="number">0</span>], L[<span class="number">1</span>]]</span><br><span class="line">        <span class="keyword">else</span>:</span><br><span class="line">            <span class="keyword">return</span> [L[<span class="number">1</span>], L[<span class="number">0</span>]]</span><br><span class="line">    <span class="keyword">else</span>:</span><br><span class="line">        middle = <span class="built_in">len</span>(L)//<span class="number">2</span></span><br><span class="line">        left = mergeSort(L[:middle])</span><br><span class="line">        right = mergeSort(L[middle:])</span><br><span class="line">        <span class="keyword">return</span> merge(left, right)</span><br><span class="line">       </span><br><span class="line">       </span><br><span class="line"><span class="comment"># 合并函数</span></span><br><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">merge</span>(<span class="params">left, right</span>):</span></span><br><span class="line">    result = []</span><br><span class="line">    i,j = <span class="number">0</span>,<span class="number">0</span></span><br><span class="line">    <span class="keyword">while</span> i &lt; <span class="built_in">len</span>(left) <span class="keyword">and</span> j &lt; <span class="built_in">len</span>(right):</span><br><span class="line">        <span class="keyword">if</span> left[i] &lt; right[j]:</span><br><span class="line">            result.append(left[i])</span><br><span class="line">            i += <span class="number">1</span></span><br><span class="line">        <span class="keyword">else</span>:</span><br><span class="line">            result.append(right[j])</span><br><span class="line">            j += <span class="number">1</span></span><br><span class="line">    <span class="keyword">while</span> i &lt; <span class="built_in">len</span>(left):</span><br><span class="line">        result.append(left[i])</span><br><span class="line">        i += <span class="number">1</span></span><br><span class="line">    <span class="keyword">while</span> j &lt; <span class="built_in">len</span>(right):</span><br><span class="line">        result.append(right[j])</span><br><span class="line">        j += <span class="number">1</span></span><br><span class="line">    <span class="keyword">return</span> result</span><br><span class="line"></span><br><span class="line"></span><br><span class="line"><span class="keyword">if</span> __name__ == <span class="string">&quot;__main__&quot;</span>:</span><br><span class="line">    L = [<span class="number">2</span>, <span class="number">4</span>, <span class="number">3</span>, <span class="number">9</span>, <span class="number">7</span>, <span class="number">8</span>, <span class="number">6</span>, <span class="number">4</span>, <span class="number">1</span>, <span class="number">5</span>, <span class="number">7</span>, <span class="number">3</span>]</span><br><span class="line">    res = mergeSort(L)</span><br><span class="line">    print(L, <span class="string">&quot;\n&quot;</span>, res)</span><br></pre></td></tr></table></figure>
<p>归并排序要比选择排序高效很多，其时间复杂度为nlog2n.<br>现在来用分治法解决铺瓷砖的问题。<br>首先看n=1的情况，那是2×2的地方，L形瓷砖可以占满3个，还剩一个。<br><img src="https://zymblog-1258069789.cos.ap-chengdu.myqcloud.com/blog0178-QTLearn/39/02.png"><br>这种情况就是我们分治法递归时的基本情况。<br>可以这么放第一个<br><img src="https://zymblog-1258069789.cos.ap-chengdu.myqcloud.com/blog0178-QTLearn/39/03.png"><br>然后就缩小为4个更小规模的问题了。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># 用递归法铺砖</span></span><br><span class="line">EMPTYPIECE = -<span class="number">1</span></span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">recursiveTile</span>(<span class="params">yard, size, originR, originC, rMiss, cMiss, nextPiece</span>):</span></span><br><span class="line">    quadMiss = <span class="number">2</span>*(rMiss &gt;= size // <span class="number">2</span>) + (cMiss &gt;= size // <span class="number">2</span>)</span><br><span class="line">    <span class="keyword">if</span> size == <span class="number">2</span>:</span><br><span class="line">        piecePos = [(<span class="number">0</span>,<span class="number">0</span>), (<span class="number">0</span>,<span class="number">1</span>), (<span class="number">1</span>,<span class="number">0</span>), (<span class="number">1</span>,<span class="number">1</span>)]</span><br><span class="line">        piecePos.pop(quadMiss)</span><br><span class="line">        <span class="keyword">for</span> (r, c) <span class="keyword">in</span> piecePos:</span><br><span class="line">            yard[originR+r][originC+c] = nextPiece</span><br><span class="line">        nextPiece = nextPiece + <span class="number">1</span></span><br><span class="line">        <span class="keyword">return</span> nextPiece</span><br><span class="line">       </span><br><span class="line">    <span class="keyword">for</span> quad <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">4</span>):</span><br><span class="line">        shiftR = size//<span class="number">2</span>*(quad &gt;= <span class="number">2</span>)</span><br><span class="line">        shiftC = size//<span class="number">2</span>*(quad % <span class="number">2</span> == <span class="number">1</span>)</span><br><span class="line">        <span class="keyword">if</span> quad == quadMiss:</span><br><span class="line">            nextPiece = recursiveTile(yard, size//<span class="number">2</span>, originR+shiftR, originC+shiftC, rMiss-shiftR, cMiss-shiftC, nextPiece)</span><br><span class="line">        <span class="keyword">else</span>:</span><br><span class="line">            newrMiss = (size//<span class="number">2</span>-<span class="number">1</span>)*(quad&lt;<span class="number">2</span>)</span><br><span class="line">            newcMiss = (size//<span class="number">2</span>-<span class="number">1</span>)*(quad%<span class="number">2</span> == <span class="number">0</span>)</span><br><span class="line">            nextPiece = recursiveTile(yard, size//<span class="number">2</span>, originR+shiftR, originC+shiftC, newrMiss, newcMiss, nextPiece)</span><br><span class="line">        centerPos = [(r + size//<span class="number">2</span> - <span class="number">1</span>, c + size//<span class="number">2</span> - <span class="number">1</span>) <span class="keyword">for</span> (r,c) <span class="keyword">in</span> [(<span class="number">0</span>,<span class="number">0</span>), (<span class="number">0</span>,<span class="number">1</span>), (<span class="number">1</span>,<span class="number">0</span>), (<span class="number">1</span>,<span class="number">1</span>)]]</span><br><span class="line">    centerPos.pop(quadMiss)</span><br><span class="line">    <span class="keyword">for</span> (r,c) <span class="keyword">in</span> centerPos:</span><br><span class="line">        yard[originR + r][originC + c] = nextPiece</span><br><span class="line">    nextPiece = nextPiece + <span class="number">1</span></span><br><span class="line"></span><br><span class="line">    <span class="keyword">return</span> nextPiece</span><br><span class="line">   </span><br><span class="line">   </span><br><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">tileMissingYard</span>(<span class="params">n, rMiss, cMiss</span>):</span></span><br><span class="line">    yard = [[EMPTYPIECE <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">2</span>**n)] <span class="keyword">for</span> j <span class="keyword">in</span> <span class="built_in">range</span>(<span class="number">2</span>**n)]</span><br><span class="line">    recursiveTile(yard, <span class="number">2</span>**n, <span class="number">0</span>, <span class="number">0</span>, rMiss, cMiss, <span class="number">0</span>)</span><br><span class="line">    <span class="keyword">return</span> yard</span><br><span class="line">   </span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">printYard</span>(<span class="params">yard</span>):</span></span><br><span class="line">    <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="built_in">len</span>(yard)):</span><br><span class="line">        row = <span class="string">&quot;&quot;</span></span><br><span class="line">        <span class="keyword">for</span> j <span class="keyword">in</span> <span class="built_in">range</span>(<span class="built_in">len</span>(yard[<span class="number">0</span>])):</span><br><span class="line">            <span class="keyword">if</span> yard[i][j] != EMPTYPIECE:</span><br><span class="line">                row += <span class="built_in">chr</span>((yard[i][j] % <span class="number">26</span>) + <span class="built_in">ord</span>(<span class="string">&quot;A&quot;</span>))</span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                row += <span class="string">&quot; &quot;</span></span><br><span class="line">        print(row)</span><br><span class="line"></span><br><span class="line">printYard(tileMissingYard(<span class="number">3</span>,<span class="number">4</span>,<span class="number">6</span>))</span><br></pre></td></tr></table></figure>
<p>说实话没太懂。原理懂了，就是分治，缩小问题规模。但具体代码……先这样吧。<br>本章代码： <a target="_blank" rel="noopener" href="https://github.com/zwdnet/MyQuant/tree/master/44">https://github.com/zwdnet/MyQuant/tree/master/44</a></p>
<p>我发文章的三个地方，欢迎大家在朋友圈等地方分享，欢迎点“在看”。<br>我的个人博客地址：<a href="https://zwdnet.github.io/">https://zwdnet.github.io</a><br>我的知乎文章地址： <a target="_blank" rel="noopener" href="https://www.zhihu.com/people/zhao-you-min/posts">https://www.zhihu.com/people/zhao-you-min/posts</a><br>我的微信个人订阅号：赵瑜敏的口腔医学学习园地</p>
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